Determining the principal value of $\frac{1}{\lvert\vec{x}-\vec{x}'\rvert^3}$

Question

I use the following definition of principal value: \begin{equation} \mathcal{P}\left(\frac{1}{f}\right)\left[h\right]=\lim_{\epsilon\to0}\sum_{x_i|f(x_i)=0}\left[\int_{-\infty}^{x_i-\epsilon}dx\frac{h(x)}{f(x)}+\int_{x_i+\epsilon}^{+\infty}dx\frac{h(x)}{f(x)}\right] \end{equation} where $h$ is a test function. With this definition I would like to compute \begin{equation} \mathcal{P}\left(\frac{1}{\lvert\vec{x}-\vec{x}'\rvert^3}\right) \end{equation} I am not sure how to compute it by using that definition. The ultimate goal for me is to compute this integral: \begin{equation} \int d^3x'\sqrt{g}\,\mathcal{P}\left(\frac{1}{\lvert\vec{x}-\vec{x}'\rvert^3}\right)\left(\frac{1}{\lvert \vec{x}'\rvert^n}\right) \end{equation} where $n$ is a non negative integer and $g$ is the metric for a 3 dimensional flate space. I could imagine to use spherical coordinates. Then I would have the integral: \begin{equation} 2\pi\int dr'd\theta'\,r'^2\sin\theta'\mathcal{P}\left(\frac{1}{(r^2+r'^2-2rr'\cos\theta)^{3/2}}\right)\frac{1}{r^{'n}} \end{equation} but I don't know how to continue. Any help is appreciated.